Business Calculus Final Exam Review: Step-by-Step Guidance

Q1. Find the area under the curve $y = x^3$ over the interval $[1,5]$.

Background

Topic: Definite Integrals and Area Under a Curve

This question tests your ability to use integration to find the area under a polynomial function between two points. This is a fundamental skill in Business Calculus, especially for applications involving total change or accumulation.

Key Terms and Formulas

• Definite Integral: $\int_a^b f(x)\,dx$ gives the area under $f(x)$ from $x=a$ to $x=b$.

• Power Rule for Integration: $\int x^n\,dx = \frac{x^{n+1}}{n+1} + C$ for $n \neq -1$.

Step-by-Step Guidance

1. Set up the definite integral for the area: $\int_1^5 x^3\,dx$.

2. Apply the power rule for integration to $x^3$: $\int x^3\,dx = \frac{x^4}{4}$.

3. Evaluate the antiderivative at the upper and lower bounds: $\left.\frac{x^4}{4}\right|_1^5$.

4. Subtract the value at $x=1$ from the value at $x=5$ to find the area.

Try solving on your own before revealing the answer!

Final Answer: 154

$\int_1^5 x^3\,dx = \left.\frac{x^4}{4}\right|_1^5 = \frac{5^4}{4} - \frac{1^4}{4} = \frac{625}{4} - \frac{1}{4} = \frac{624}{4} = 156$

The area under the curve $y = x^3$ from $x=1$ to $x=5$ is 156.

Q2. Find the area under the curve $y = x^4$ over the interval $[1,3]$.

Background

Topic: Definite Integrals and Area Under a Curve

This question tests your ability to use integration to find the area under a higher-degree polynomial function between two points. This is important for understanding accumulation and total change in business applications.

Key Terms and Formulas

• Definite Integral: $\int_a^b f(x)\,dx$ gives the area under $f(x)$ from $x=a$ to $x=b$.

• Power Rule for Integration: $\int x^n\,dx = \frac{x^{n+1}}{n+1} + C$ for $n \neq -1$.

Step-by-Step Guidance

1. Set up the definite integral for the area: $\int_1^3 x^4\,dx$.

2. Apply the power rule for integration to $x^4$: $\int x^4\,dx = \frac{x^5}{5}$.

3. Evaluate the antiderivative at the upper and lower bounds: $\left.\frac{x^5}{5}\right|_1^3$.

4. Subtract the value at $x=1$ from the value at $x=3$ to find the area.

Try solving on your own before revealing the answer!

Final Answer: 48

$\int_1^3 x^4\,dx = \left.\frac{x^5}{5}\right|_1^3 = \frac{243}{5} - \frac{1}{5} = \frac{242}{5} = 48.4$

The area under the curve $y = x^4$ from $x=1$ to $x=3$ is 48.4.